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Alex787 [66]
3 years ago
7

Prepare 10.00 mL of 0.010 M NaOH by diluting the NaOH solution used in Trial 1. (Your procedure should clearly show your calcula

tions and the glassware used to perform the dilution.)
Chemistry
1 answer:
denis-greek [22]3 years ago
3 0

Answer:

<em> = 0.2 mL.</em>

Explanation:

Given a 0.5 M solution of NaOH as stock solution, 10.0mL of 0.010M can be prepared via dilution with distilled water, by using the formula:

C_{1} V_{1} = C_{2} V_{2}

where C1 and V1 are initial concentration and volume respectively; same as C2 & V2 for fina.

Let C1 = 0.5M, V2 = ?

C2 = 0.010M; V2 = 10mL

⇒Volume of stock solution to be diluted, V2

= \frac{10}{0.5} × 0.010

<em> = 0.2 mL.</em>

Glasswares used would be pipette (for smaller volume experiment) and measuring cylinder. 0.2mL would be measured and then made upto the 10mL mark of the measuring cylinder.

I hope this was a detailed explanation given the missing details of "Trial 1" in the question.

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Data Given:
                  Pressure  =  P  =  ?

                  Volume  =  V  =  3.0 L

                  Temperature  =  T  =  115 °C + 273  =  388 K

                  Mass  =  m  =  75.0 g

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Solution:
              Let suppose the Gas is acting Ideally. Then According to Ideal Gas Equation,
                                      P V  =  n R T
Solving for P,
                                      P  =  n R T / V      ------ (1)
Calculating Moles,
                                      n  =  m / M

                                      n  =  75.0 g / 44 g.mol⁻¹

                                      n  =  1.704 mol

Putting Values in Eq. 1,

                    P  =  (1.704 mol × 0.08205 atm.L.mol⁻¹.K⁻¹ × 388 K) ÷ 3.0 L

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3 years ago
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Answer:

2M

Explanation:

M=mol/L

1. Find moles of CoCl2

mass of substance/molar mass = 130/129.833 = 1.001 mol

3. Substitute in molarity equation

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3 years ago
A 4.5 liter sample of a gas at 2 atm and 300 K has 0.80 mole of gas. If 0.35 mole of the gas is added to the sample at the same
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Answer:

6.5 liters​.

Explanation:

  • We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R  is the general gas constant,

T is the temperature of the gas in K.

  • If P and T are constant, and have different values of n and V:

<em>(V₁n₂) = (V₂n₁)</em>

<em></em>

  • Knowing that:

V₁  = 4.5 L, n₁ = 0.80 mol,

V₂  =  ??? L, n₂ = 0.35 + 0.80 = 1.15 mol.

  • Applying in the above equation

<em>(V₁n₂) = (V₂n₁)</em>

<em></em>

<em>∴ V₂ = (V₁n₂)/n₁ </em>= (4.5 L)(1.15 mol)/(0.8 mol) = <em>6.469 L ≅ 6.5 L.</em>

<em></em>

<em>So, the right choice is: 6.5 liters​.</em>

6 0
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